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NeTakaya
3 years ago
12

A 70.0-kg person throws a 0.0420-kg snowball forward with a ground speed of 35.0 m/s. A second person, with a mass of 57.0 kg, c

atches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.20 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged?
Physics
1 answer:
ollegr [7]3 years ago
3 0

Answer:

so throwers new velocity = 2.18032m/s

so catchers new velocity = 0.02577m/s

Explanation:

Directly by conservation of momentum we can write

m_1u_1+m_2u_2= m_1v_1+m_2v_2

let x be the thrower's new velocity

(70+0.042)×2.2 + 57×0 = 70× x +0.042×35 +57×0

x = 2.18032m/s

so the velocity of 70 kg man = 2.18032m/s

so throwers new velocity = 2.18032m/s

now again by conservation of momentum

0.042×35 = (57+0.042) ×y

y = 0.02577m/s

so catchers new velocity = 0.02577m/s

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Answer: Interneurons.

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At a certain location, a gravitational force with a magnitude of 350 newtons acts on a 70.-kilogram astronaut. What is the magni
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Answer:

3. 5.0N/kg

Explanation:

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Means/end analysis involves:_________
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a. a gradual approximation to the final solution

Explanation:

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4 0
3 years ago
An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equil
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Answer:

The amount of ice added in gram is 32.77g

Explanation:

This problem bothers on the heat capacity of materials

Given data

Mass of water Mw= 200g

Temperature of water θw= 25°c

Temperature of ice θice= 0°c

Equilibrium Temperature θe= 12°c

Mass of ice Mi=???

The specific heat of ice Ci= 2090 J/(kg ∘C)

specific heat of water Cw = 4186 J/(kg ∘C)

latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

heat heat loss by water = heat gained by ice

N/B let us understand something, heat gained by ice is in two phases

Heat require to melt ice at 0°C to water at 0°C

And the heat required to take water from 0°C to equilibrium temperature

Hence

MwCwΔθ=MiLi +MiCiΔθ

Substituting our data we have

200*4186*(25-12)=Mi*3.3x10^5+

Mi*2090(12-0)

837200*13=Mi*3.3x10^5+Mi*2090

10883600=332090Mi

Mi=10883600/332090

Mi= 32.77g

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3 years ago
A force of 500 N is exerted on a baseball by the bat for 0.001 s. What is the change in momentum of the baseball?
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3 years ago
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